I call this function with onSubmit = "return ajaxreq(this,"myfile.php");" in my <form> tag so that looks legit.

GJay

06-07-2007, 07:19 PM

you're going to need to show more code, it's not clear what 'this' is in the code you've pasted.
Your explanation mentions 'this.elements', but the code has thisForm.elements. That's a little confusing, more code should clarify.